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AP Calculus BC Flashcards: Calculating Higher Order Derivatives

Study Calculating Higher Order Derivatives in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Calculating Higher Order Derivatives, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Calculating Higher Order Derivatives

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QUESTION

Find the second derivative of f(x)=e−xf(x) = e^{-x}f(x)=e−x.

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ANSWER

f′′(x)=e−xf''(x) = e^{-x}f′′(x)=e−x. Derivative of e−xe^{-x}e−x multiplies by (−1)2=1(-1)^2 = 1(−1)2=1.

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Flashcard 1: Find the second derivative of f(x)=e−xf(x) = e^{-x}f(x)=e−x.

Answer: f′′(x)=e−xf''(x) = e^{-x}f′′(x)=e−x. Derivative of e−xe^{-x}e−x multiplies by (−1)2=1(-1)^2 = 1(−1)2=1.

Flashcard 2: What is the fourth derivative of f(x)=x4−x2+1f(x) = x^4 - x^2 + 1f(x)=x4−x2+1?

Answer: f(4)(x)=24f^{(4)}(x) = 24f(4)(x)=24. Fourth derivative eliminates lower-order terms, leaving only x4x^4x4 coefficient.

Flashcard 3: State the second derivative of f(x)=tan(x)f(x) = \text{tan}(x)f(x)=tan(x).

Answer: f′′(x)=2sec2(x)tan(x)f''(x) = 2\text{sec}^2(x)\text{tan}(x)f′′(x)=2sec2(x)tan(x). First derivative is sec⁡2(x)\sec^2(x)sec2(x), apply chain rule again.

Flashcard 4: State the formula for the second derivative of f(x)=ln(x)f(x) = \text{ln}(x)f(x)=ln(x).

Answer: f′′(x)=−1x2f''(x) = -\frac{1}{x^2}f′′(x)=−x21​. Standard formula for logarithmic second derivative.

Flashcard 5: What is the second derivative of f(x)=e2xf(x) = \text{e}^{2x}f(x)=e2x?

Answer: f′′(x)=4e2xf''(x) = 4\text{e}^{2x}f′′(x)=4e2x. Chain rule with exponential: derivative multiplies by the inner function coefficient squared.

Flashcard 6: Compute the third derivative of f(x)=1x2f(x) = \frac{1}{x^2}f(x)=x21​.

Answer: f′′′(x)=6x5f'''(x) = \frac{6}{x^5}f′′′(x)=x56​. Rewrite as x−2x^{-2}x−2 and apply power rule three times.

Flashcard 7: Compute the third derivative of f(x)=e−2xf(x) = \text{e}^{-2x}f(x)=e−2x.

Answer: f′′′(x)=−8e−2xf'''(x) = -8\text{e}^{-2x}f′′′(x)=−8e−2x. Chain rule with e−2xe^{-2x}e−2x: coefficient becomes (−2)3=−8(-2)^3 = -8(−2)3=−8.

Flashcard 8: What is the second derivative of f(x)=arcsin(x)f(x) = \text{arcsin}(x)f(x)=arcsin(x)?

Answer: f′′(x)=x(1−x2)3/2f''(x) = \frac{x}{(1-x^2)^{3/2}}f′′(x)=(1−x2)3/2x​. Inverse trig derivatives involve radical expressions in denominators.

Flashcard 9: What is the second derivative of f(x)=xln(x)f(x) = x \text{ln}(x)f(x)=xln(x)?

Answer: f′′(x)=1xf''(x) = \frac{1}{x}f′′(x)=x1​. Use product rule: f′(x)=ln⁡(x)+1f'(x) = \ln(x) + 1f′(x)=ln(x)+1, then f′′(x)=1xf''(x) = \frac{1}{x}f′′(x)=x1​.

Flashcard 10: What is the second derivative of f(x)=x2exf(x) = x^2 \text{e}^xf(x)=x2ex?

Answer: f′′(x)=(x2+4x+2)exf''(x) = (x^2 + 4x + 2)\text{e}^xf′′(x)=(x2+4x+2)ex. Product rule applied twice to x2exx^2 e^xx2ex.

Flashcard 11: Compute the fourth derivative of f(x)=cos(3x)f(x) = \text{cos}(3x)f(x)=cos(3x).

Answer: f(4)(x)=81cos(3x)f^{(4)}(x) = 81\text{cos}(3x)f(4)(x)=81cos(3x). Fourth derivative of cos⁡(3x)\cos(3x)cos(3x) involves 34=813^4 = 8134=81 and returns to cosine.

Flashcard 12: Find the third derivative of f(x)=5x4+3x3−xf(x) = 5x^4 + 3x^3 - xf(x)=5x4+3x3−x.

Answer: f′′′(x)=120x+18f'''(x) = 120x + 18f′′′(x)=120x+18. Apply power rule to each term and differentiate three times.

Flashcard 13: Compute the third derivative of f(x)=sin(x)f(x) = \text{sin}(x)f(x)=sin(x).

Answer: f′′′(x)=−cos(x)f'''(x) = -\text{cos}(x)f′′′(x)=−cos(x). Trig derivatives cycle: sin⁡→cos⁡→−sin⁡→−cos⁡\sin \to \cos \to -\sin \to -\cossin→cos→−sin→−cos.

Flashcard 14: What is the third derivative of f(x)=sin(2x)f(x) = \text{sin}(2x)f(x)=sin(2x)?

Answer: f′′′(x)=−8sin(2x)f'''(x) = -8\text{sin}(2x)f′′′(x)=−8sin(2x). Chain rule with sin⁡(2x)\sin(2x)sin(2x): coefficient becomes (−2)3=−8(-2)^3 = -8(−2)3=−8.

Flashcard 15: What is the second derivative of f(x)=e3xf(x) = \text{e}^{3x}f(x)=e3x?

Answer: f′′(x)=9e3xf''(x) = 9\text{e}^{3x}f′′(x)=9e3x. Chain rule with e3xe^{3x}e3x: coefficient becomes 32=93^2 = 932=9.

Flashcard 16: Find the fourth derivative of f(x)=x4−3x2+2f(x) = x^4 - 3x^2 + 2f(x)=x4−3x2+2.

Answer: f(4)(x)=24f^{(4)}(x) = 24f(4)(x)=24. Fourth derivative of x4x^4x4 term is 242424, other terms become zero.

Flashcard 17: Find the third derivative of f(x)=13x3f(x) = \frac{1}{3}x^3f(x)=31​x3.

Answer: f′′′(x)=2f'''(x) = 2f′′′(x)=2. Third derivative eliminates the coefficient 13\frac{1}{3}31​ leaving constant 222.

Flashcard 18: What is the second derivative of f(x)=1xf(x) = \frac{1}{x}f(x)=x1​?

Answer: f′′(x)=2x3f''(x) = \frac{2}{x^3}f′′(x)=x32​. Rewrite as x−1x^{-1}x−1, then apply power rule twice.

Flashcard 19: What is the fourth derivative of f(x)=x4f(x) = x^4f(x)=x4?

Answer: f(4)(x)=24f^{(4)}(x) = 24f(4)(x)=24. Fourth derivative of x4x^4x4 gives the factorial 4!=244! = 244!=24.

Flashcard 20: State the second derivative of f(x)=cos(x)f(x) = \text{cos}(x)f(x)=cos(x).

Answer: f′′(x)=−cos(x)f''(x) = -\text{cos}(x)f′′(x)=−cos(x). Second derivative of cos⁡(x)\cos(x)cos(x) follows trig cycle pattern.

Flashcard 21: What is the second derivative of f(x)=exf(x) = e^xf(x)=ex?

Answer: f′′(x)=exf''(x) = e^xf′′(x)=ex. Derivative of exe^xex is always exe^xex.

Flashcard 22: What is the second derivative of f(x)=sin2(x)f(x) = \text{sin}^2(x)f(x)=sin2(x)?

Answer: f′′(x)=2cos(2x)f''(x) = 2\text{cos}(2x)f′′(x)=2cos(2x). Use double angle identity: sin⁡2(x)=1−cos⁡(2x)2\sin^2(x) = \frac{1-\cos(2x)}{2}sin2(x)=21−cos(2x)​.

Flashcard 23: Compute the third derivative of f(x)=x2sin(x)f(x) = x^2 \text{sin}(x)f(x)=x2sin(x).

Answer: f′′′(x)=−2(3cos(x)+xsin(x))f'''(x) = -2(3\text{cos}(x) + x\text{sin}(x))f′′′(x)=−2(3cos(x)+xsin(x)). Use product rule repeatedly on x2sin⁡(x)x^2 \sin(x)x2sin(x).

Flashcard 24: Compute the third derivative of f(x)=e−2xf(x) = \text{e}^{-2x}f(x)=e−2x.

Answer: f′′′(x)=−8e−2xf'''(x) = -8\text{e}^{-2x}f′′′(x)=−8e−2x. Chain rule with e−2xe^{-2x}e−2x: coefficient becomes (−2)3=−8(-2)^3 = -8(−2)3=−8.

Flashcard 25: State the formula for the second derivative of f(x)=ln(x)f(x) = \text{ln}(x)f(x)=ln(x).

Answer: f′′(x)=−1x2f''(x) = -\frac{1}{x^2}f′′(x)=−x21​. Standard formula for logarithmic second derivative.

Flashcard 26: What is the second derivative of f(x)=ln(x2)f(x) = \text{ln}(x^2)f(x)=ln(x2)?

Answer: f′′(x)=−2x2f''(x) = -\frac{2}{x^2}f′′(x)=−x22​. Use chain rule: ln⁡(x2)=2ln⁡(x)\ln(x^2) = 2\ln(x)ln(x2)=2ln(x), so derivative doubles.

Flashcard 27: Find the third derivative of f(x)=5x4+3x3−xf(x) = 5x^4 + 3x^3 - xf(x)=5x4+3x3−x.

Answer: f′′′(x)=120x+18f'''(x) = 120x + 18f′′′(x)=120x+18. Apply power rule to each term and differentiate three times.

Flashcard 28: What is the fourth derivative of f(x)=x4−x2+1f(x) = x^4 - x^2 + 1f(x)=x4−x2+1?

Answer: f(4)(x)=24f^{(4)}(x) = 24f(4)(x)=24. Fourth derivative eliminates lower-order terms, leaving only x4x^4x4 coefficient.

Flashcard 29: What is the second derivative of f(x)=e3xf(x) = \text{e}^{3x}f(x)=e3x?

Answer: f′′(x)=9e3xf''(x) = 9\text{e}^{3x}f′′(x)=9e3x. Chain rule with e3xe^{3x}e3x: coefficient becomes 32=93^2 = 932=9.

Flashcard 30: State the second derivative of f(x)=tan(x)f(x) = \text{tan}(x)f(x)=tan(x).

Answer: f′′(x)=2sec2(x)tan(x)f''(x) = 2\text{sec}^2(x)\text{tan}(x)f′′(x)=2sec2(x)tan(x). First derivative is sec⁡2(x)\sec^2(x)sec2(x), apply chain rule again.